Question

The sum of coefficients of all the integral powers of  $$x$$ in the expansion of $$(1+2\sqrt{x})^{40}$$ is

A
340+1
B
3401
C
12 (3401)
D
12 (340+1)

Solution

The correct option is D $$\displaystyle\frac{1}{2}$$ $$(3^{40}+1)$$$$(1+2\sqrt{x})^{40}=1+\:^{40}C_{1}(2\sqrt{x})^{1}+\:^{40}C_{2}(2\sqrt{x})^{2}+\:^{40}C_{3}(2\sqrt{x})^{3}+...+\:^{40}C_{40}(2\sqrt{x})^{40}$$Hence, all the odd terms will have integral powers of $$x$$.Substituting $$x=1$$, we get$$3^{40}=a_{0}+a_{1}+a_{2}+a_{3}...+a_{40}$$ ...(i)Now consider,$$(1-2\sqrt{x})^{40}=1-\:^{40}C_{1}(2\sqrt{x})^{1}+\:^{40}C_{2}(2\sqrt{x})^{2}-\:^{40}C_{3}(2\sqrt{x})^{3}...+\:^{40}C_{40}(2\sqrt{x})^{40}$$Taking $$x$$ as $$1$$, we get$$1=a_{0}-a_{1}+a_{2}-a_{3}...+a_{40}$$ ...(ii)Adding i and, ii we get$$3^{40}+1=2[a_{0}+a_{2}+a_{4}+a_{6}...+a_{40}]$$$$\dfrac{3^{40}+1}{2}=a_{0}+a_{2}+a_{4}+a_{6}...+a_{40}$$Hence the sum of the coefficients, of integral powers of $$x$$ will be$$\dfrac{3^{40}+1}{2}$$Hence, option D is correctMathematics

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